Convergence of generic p-series
0
$begingroup$
Consider the generic p-series:
$sum_{n=1}^{infty}frac{1}{n^{p}}$
Let the increment function be: $n= n+K$
where $K$ is some integer constant. For $K=1$ its simple p-series and that diverges for any $p<1$.
Can we have some integer value of $K$ beyond which the series converges for $p<1$. Specifically for $ 1/2 < p <1.$
real-analysis
asked Dec 18 '18 at 10:27
TheoryQuest1TheoryQuest1
1417
$endgroup$
add a comment |
0
$begingroup$
Consider the generic p-series:
$sum_{n=1}^{infty}frac{1}{n^{p}}$
Let the increment function be: $n= n+K$
where $K$ is some integer constant. For $K=1$ its simple p-series and that diverges for any $p<1$.
Can we have some integer value of $K$ beyond which the series converges for $p<1$. Specifically for $ 1/2 < p <1.$
real-analysis
asked Dec 18 '18 at 10:27
TheoryQuest1TheoryQuest1
1417
$endgroup$
1
$begingroup$
$displaystylesum_{n=1}^{infty}frac{1}{(n+K)^p}$ converges iff $p>1$ (regardless of $K$). Did you mean something else?
$endgroup$
– metamorphy
Dec 18 '18 at 10:57
$begingroup$
Ah it didn't sorry. The question is what if we consider every $K^{th}$ term for the values of $n$. Does the same constraint still hold true? or can we for some $K$ have the series converge even if $p<1$. I would thus rewrite the question as $displaystylesum_{n=1}^{infty}frac{1}{(n*K)^p}$
$endgroup$
– TheoryQuest1
Dec 18 '18 at 11:13
1
$begingroup$
Then it is $displaystylefrac{1}{K^p}sum_{n=1}^{infty}frac{1}{n^p}$. With the same outcome.
$endgroup$
– metamorphy
Dec 18 '18 at 11:16
add a comment |
0
0
0
$begingroup$
Consider the generic p-series:
$sum_{n=1}^{infty}frac{1}{n^{p}}$
Let the increment function be: $n= n+K$
where $K$ is some integer constant. For $K=1$ its simple p-series and that diverges for any $p<1$.
Can we have some integer value of $K$ beyond which the series converges for $p<1$. Specifically for $ 1/2 < p <1.$
real-analysis
asked Dec 18 '18 at 10:27
TheoryQuest1TheoryQuest1
1417
$endgroup$
Consider the generic p-series:
$sum_{n=1}^{infty}frac{1}{n^{p}}$
Let the increment function be: $n= n+K$
where $K$ is some integer constant. For $K=1$ its simple p-series and that diverges for any $p<1$.
Can we have some integer value of $K$ beyond which the series converges for $p<1$. Specifically for $ 1/2 < p <1.$
real-analysis
real-analysis
asked Dec 18 '18 at 10:27
TheoryQuest1TheoryQuest1
1417
asked Dec 18 '18 at 10:27
TheoryQuest1TheoryQuest1
1417
asked Dec 18 '18 at 10:27
TheoryQuest1TheoryQuest1
1417
asked Dec 18 '18 at 10:27
TheoryQuest1TheoryQuest1
1417
asked Dec 18 '18 at 10:27
TheoryQuest1TheoryQuest1
1417
1417
1
$begingroup$
$displaystylesum_{n=1}^{infty}frac{1}{(n+K)^p}$ converges iff $p>1$ (regardless of $K$). Did you mean something else?
$endgroup$
– metamorphy
Dec 18 '18 at 10:57
$begingroup$
Ah it didn't sorry. The question is what if we consider every $K^{th}$ term for the values of $n$. Does the same constraint still hold true? or can we for some $K$ have the series converge even if $p<1$. I would thus rewrite the question as $displaystylesum_{n=1}^{infty}frac{1}{(n*K)^p}$
$endgroup$
– TheoryQuest1
Dec 18 '18 at 11:13
1
$begingroup$
Then it is $displaystylefrac{1}{K^p}sum_{n=1}^{infty}frac{1}{n^p}$. With the same outcome.
$endgroup$
– metamorphy
Dec 18 '18 at 11:16
add a comment |
1
$begingroup$
$displaystylesum_{n=1}^{infty}frac{1}{(n+K)^p}$ converges iff $p>1$ (regardless of $K$). Did you mean something else?
$endgroup$
– metamorphy
Dec 18 '18 at 10:57
$begingroup$
Ah it didn't sorry. The question is what if we consider every $K^{th}$ term for the values of $n$. Does the same constraint still hold true? or can we for some $K$ have the series converge even if $p<1$. I would thus rewrite the question as $displaystylesum_{n=1}^{infty}frac{1}{(n*K)^p}$
$endgroup$
– TheoryQuest1
Dec 18 '18 at 11:13
1
$begingroup$
Then it is $displaystylefrac{1}{K^p}sum_{n=1}^{infty}frac{1}{n^p}$. With the same outcome.
$endgroup$
– metamorphy
Dec 18 '18 at 11:16
1
1
$begingroup$
$displaystylesum_{n=1}^{infty}frac{1}{(n+K)^p}$ converges iff $p>1$ (regardless of $K$). Did you mean something else?
$endgroup$
– metamorphy
Dec 18 '18 at 10:57
$begingroup$
$displaystylesum_{n=1}^{infty}frac{1}{(n+K)^p}$ converges iff $p>1$ (regardless of $K$). Did you mean something else?
$endgroup$
– metamorphy
Dec 18 '18 at 10:57
$begingroup$
Ah it didn't sorry. The question is what if we consider every $K^{th}$ term for the values of $n$. Does the same constraint still hold true? or can we for some $K$ have the series converge even if $p<1$. I would thus rewrite the question as $displaystylesum_{n=1}^{infty}frac{1}{(n*K)^p}$
$endgroup$
– TheoryQuest1
Dec 18 '18 at 11:13
$begingroup$
Ah it didn't sorry. The question is what if we consider every $K^{th}$ term for the values of $n$. Does the same constraint still hold true? or can we for some $K$ have the series converge even if $p<1$. I would thus rewrite the question as $displaystylesum_{n=1}^{infty}frac{1}{(n*K)^p}$
$endgroup$
– TheoryQuest1
Dec 18 '18 at 11:13
1
1
$begingroup$
Then it is $displaystylefrac{1}{K^p}sum_{n=1}^{infty}frac{1}{n^p}$. With the same outcome.
$endgroup$
– metamorphy
Dec 18 '18 at 11:16
$begingroup$
Then it is $displaystylefrac{1}{K^p}sum_{n=1}^{infty}frac{1}{n^p}$. With the same outcome.
$endgroup$
– metamorphy
Dec 18 '18 at 11:16
add a comment |
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1
$begingroup$
$displaystylesum_{n=1}^{infty}frac{1}{(n+K)^p}$ converges iff $p>1$ (regardless of $K$). Did you mean something else?
$endgroup$
– metamorphy
Dec 18 '18 at 10:57
$begingroup$
Ah it didn't sorry. The question is what if we consider every $K^{th}$ term for the values of $n$. Does the same constraint still hold true? or can we for some $K$ have the series converge even if $p<1$. I would thus rewrite the question as $displaystylesum_{n=1}^{infty}frac{1}{(n*K)^p}$
$endgroup$
– TheoryQuest1
Dec 18 '18 at 11:13
1
$begingroup$
Then it is $displaystylefrac{1}{K^p}sum_{n=1}^{infty}frac{1}{n^p}$. With the same outcome.
$endgroup$
– metamorphy
Dec 18 '18 at 11:16